Chapter.3 Design of Commutator and Brushes

8/18/2019 Chapter.3 Design of Commutator and Brushes

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Chapter.3 Design of Commutator and Brushes

The Commutator is an assembly of Commutator segments or bars tapered in section. The

segments made of hard drawn copper are insulated from each other by mica or micanite, the

usual thickness of which is about 0.8 mm. The number of commutator segments is equal to thenumber of active armature coils.

The diameter of the commutator will generally be about (60 to 80)% of the armature diameter.Lesser values are used for high capacity machines and higher values for low capacity

machines.

Higher values of commutator peripheral velocity are to be avoided as it leads to lesser

commutation time dt, increased reactance voltagedi

RV = Ldt

and sparking commutation.

The commutator peripheral velocity vc = π DC N / 60 should not as for as possible be morethan about 15 m/s. (Peripheral velocity of 30 m/s is also being used in practice but should be

avoided whenever possible.)

The commutator segment pitch τC = (outside width of one segment + mica insulation between

segments) = π DC / Number of segments should not be less than 4 mm. (This minimumsegment pitch is due to 3.2 mm of copper + 0.8 mm of mica insulation between segments.) The

outer surface width of commutator segment lies between 4 and 20 mm in practice.

The axial length of the commutator depends on the space required



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1) by the brushes with brush boxes

2) for the staggering of brushes3)

for the margin between the end of commutator and brush and

4) for the margin between the brush and riser and width of riser.

If there are nb brushes / brush arm or spindle or holder, placed one beside the other on thecommutator surface, then the length of the commutator LC = (width of the brush wb + brush

box thickness 0.5 cm) number of brushes / spindle + end clearance 2 to 4 cm + clearance forrisers 2 to 4 cm + clearance for staggering of brushes 2 to 4 cm.

If the length of the commutator (as calculated from the above expression) leads to smalldissipating surface π DC LC, then the commutator length must be increased so that thetemperature rise of the commutator does not exceed a permissible value say 55

0C.

The temperature rise of the commutator can be calculated by using the following empirical

formula.

2o C C

C

120 watt loss / cm of dissipating surface D Lθ C =

1 0.1 v

π ×

+

The different losses that are responsible for the temperature rise of the commutator are (a)brush contact loss and (b) brush frictional loss.

Brush contact loss = voltage drop / brush set × Ia

The voltage drop / brush set depend on the brush material – Carbon, graphite, electro graphite

or metalized graphite. The voltage drop / brush set can be taken as 2.0 V for carbon brushes.

Brush frictional loss (due to all the brush arms)

= frictional torque in Nm × angular velocity



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= frictional force in Newton x distance in metre ×60

N2π

= 9.81 µPbAball×2

DC

/×

60

N2π /= 9.81µPbAball Cv

whereµ = coefficient of friction and depends on the brush material. Lies between 0.22 and 0.27for carbon brushes

Pb = Brush pressure in kg / m2 and lies between 1000 and 1500

Aball = Area of the brushes of all the brush arms in m2

= Ab× number of brush arms

= Ab× number of poles in case of lap winding

= Ab× 2 or P in case of wave windingAb = Cross-sectional area of the brush / brush arm

Brush Details

Since the brushes of each brush arm collets the current from two parallel paths, current

collected by each brush arm isA

I2 a and the cross-sectional area of the brush or brush arm or

holder or spindle Ab =b

a

A

I2

δ cm2. The current density bδ depends on the brush material and

can be assumed between 5.5 and 6.5 A / cm2 for carbon.

In order to ensure a continuous supply of power and cost of replacement of damaged or wornout brushes is cheaper, a number of subdivided brushes are used instead of one single brush.

Thus if

i)

tb is the thickness of the brush

ii)

wb is the width of the brush and

iii) nbis the number of sub divided brushes

thenAb = tbwbnb

As the number of adjacent coils of the same or different slots that are simultaneously undergoing commutation increases, the brush width and time of commutation also increases at the

same rate and therefore the reactance voltage (the basic cause of sparking commutation)

becomes independent of brush width.

With only one coil under going commutation and width of the brush equal to one segment

width, the reactance voltage and hence the sparking increases as the slot width decreases.Hence the brush width is made to cover more than one segment. If the brush is too wide, then

those coils which are away from the commutating pole zone or coils not coming under the

influence of inter pole flux and under going commutation leads to sparking commutation.



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Hence brush width greater than the commutating zone width is not advisable under any

circumstances. Since the commutating pole zone lies between (9 and 15)% of the pole pitch,15% of the commutator circumference can be considered as the maximum width of the brush.

It has been found that the brush width should not be more than 5 segments in machines less

than 50 kW and 4 segments in machines more than 50 kW.

The number of brushes / spindle can be found out by assuming a standard brush width or amaximum current / sub divided brush.

Standard brush width can be 1.6, 2.2 or 3.2 cm

Current/subdivided brush should not be more than 70A

Thus with the brush width assumed, nb=wt

A

bb

b . With the current / sub divided brush assumed

nb = 70A x

I2 a

and wb = nt

A

bb

b

Note :

A)

Staggering of Brushes :

Because of the current flowing from commutator segments to the brush, copper is eaten

away leading to formation of ridges between the subdivided brushes of the same brusharm. Since it is not possible to avoid eating away copper by the arc, eating away of copper

must be made to take place over the entire axial length of the commutator to ensure

uniform commutator surface. This is achieved by displacing all the positive brushes in one

direction and all the negative brushes in the other direction or by staggering of brushes inpairs as shown below.



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B) Brush materials and their properties

MaterialPeripheral

velocity m/s

Current

density in

A/cm2

Voltage drop

per brush set

in volts

Coefficient of

friction

Normal carbon 5 to 15 5.5 to 6.5 2.0 0.22 to 0.27 Soft graphite 10 to 25 9.0 to 9.5 1.6 0.12 Metalized graphite

(copper carbon mixture) 5 to 15 15 to 16 0.24 to 0.35 0.16

Electro graphite

(Graphitized by heating)

5 to 15 8.5 to 9.0 1.7 to 1.8 0.22

C) Step by step design procedure of commutator and brushes1)

Diameter of the commutator DC = (0.6 to 0.8) D and must be such that the peripheral

velocity of the commutator vC = π DC N / 60 is not more than 15 m/s as far as possible.

2)

The commutator segment pitch τC = π DC / Number of segments should not be less than 4mm from the mechanical strength point of view.

3) The number of commutator segments is equal to number of active armature coils.

4)

Length of the commutator LC = (width of the brush + brush box thickness 0.5 cm)

number of brushes / spindle nb + end clearance 2 to 4 cm + clearance for risers 2 to 4 cm +

clearance for staggering of brushes 2 to 4 cm.

5)

Cross-sectional area of the brush / spindle or arm or holder Ab =b

a

A

I2

δ cm2. The current

density in the brushes bδ lies between 5.5 and 6.5 A / cm2

for carbon brushes.

6)

Maximum thickness of the brush tb max = 4 τC for machines greater than 50 kW

= 5 τC for machines less than 50 kW

7) With standard brush width Wb assumed, the number of brushes / spindle nb

=Wt

A

bb

b

8) Total commutator losses = Brush contact loss + Brush frictional loss

= voltage drop / brush set × Ia + 9.81 µPbAball Cv

wherevoltage drop / set = 2.0 V for carbon brushes

µ = coefficient of friction and lies between 0.22 to 0.27 for carbon brushes

Pb = Brush pressure and lies between 1000 and 1500 kg / m2



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9)

Temperature rise of the commutator

θ0C = Cooling coefficient x watt loss / dissipating surface

=Cv0.11

120

+

× watt loss / cm2 of dissipating surface π DC LC

10) Temperature rise should be less than about 550C.

*******



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DESIGN OF COMMUTATOR AND BRUSHES

Example.1

A 500kW, 500V, 375 rpm, 8 pole dc generator has an armature diameter of 110 cm and the

number of armature conductor is 896. Calculate the diameter of the commutator, length of thecommutator, number of brushes per spindle, commutator losses and temperature rise of the

commutator. Assume single turn coils.

Diameter of the commutator DC = (0.6 to 0.8) D = 0.7 x 110 = 77cm

Length of the commutator LC = (width of the brush Wb + brush box thickness 0.5 cm) number

of brushes / spindle nb + end clearance 2 to 4 cm + clearance for risers 2 to 4 cm + clearance

for staggering of brushes 2 to 4 cm.

Armature current Ia =

V

10kW x 3

=

500

10 x500 3

= 1000A

Note : An armature current of 1000 A obviously calls for a lap winding.

Cross-sectional area of the brush per spindle or brush arm or holder Ab =

b

a

A

I2

δ since the current density lies between 5.5 and 6.5 A/cm

2 for carbon brushes,

let it be 6 A/mm2

Ab =6 x8

1000 x2= 41.66 cm

2

maximum thickness of the brush = 4 τC

Commutator segment pitch τC =π DC / Number of segments or coils

Number of coils = Z / 2 x number of turns per coil = 896 / 2 x 1 = 448

Therefore τC =448

77 xπ = 0.54 cm

Maximum thickness of the brush = 4 x 0.54 = 2.16 cm

Let the thickness of the brush tb= 2.0 cm

If a brush width of 2.2 cm (a standard value) is assumed then Wb = 2.2 cm

Therefore, number of brushes / spindle

nb=bb

b

Wt

A =

2.2 x2

41.66 = 9.46 and is not possible

Let the number of brushes / spindle be = 10



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Therefore LC = (2.2 + 0.5) 10 + 2 + 2 + 2 = 33 cm

Brush contact loss = voltage drop / brush set x Ia = 2 x 1000 = 2000 W

Brush frictional loss = 9.81 µPbAball Cv

Let the coefficient of friction µ = 0.25 as it lies between 0.22 to 0.27 for carbon brushes.

Let the brush pressure Pb= 1215 kg/m2 as it lies between 1000 to 1500 kg/m

2

Aball = Area of the brushes of all the brush arms

= tbwbnb x number of brush arms or number of poles as the number of brush

arms number of poles for a lap winding

= 2 x 2.2 x 10 x 8 x 10-4

= 0.0352 m2

Brush frictional loss = 9.81 x 0.25 x 1215 x 0.352 x 15.1 = 1583.8 W

Therefore commutator losses (total) = 2000 + 1583.8 = 3583.8 W

Temperature rise in degree centigrade

θ =C

CC

2

v0.1 1

LDsurface gdissipatin commutator theof cm /loss x watt120

+

π

=15.1 x0.1 1

33 x77 x /3583.8 x120

+

π = 21.46

Example.2

A 20 Hp, 4 pole, 250V, 1000 rpm wave wound D.C. machine has the following design

data. Diameter of the armature = 25 cm, number of slots = 41, number of coil sides / slot =

4, turns / coil = 2. Calculate the number of segments, outside width of one segment and

mica, brush thickness, length of the commutator and brush contact loss.

Number of segments = number of active coils for a wave winding

i) Number of coil sides (total) = 41 x 4 = 164

Number of coils = 164 / 2 = 82 as each coil will have 2 coil sides OR

ii)

Since the coil is of 2 turns, each coil side will have 2 conductors and therefore thenumber of conductors per slot = 4 x 2 = 8.

Total number of conductors = 41 x 8 = 328

Number of coils = 82 2 x2

328

coil /turnsof numberx2

conductors of numbertotal== OR

iii) Number of coils = number of slots x number of coil sides / layer

= 41 x 2 = 82



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For a wave winding YC =p

1 C ± must be an integer. With the number of coils calculated, YC

=2

1 82 ± is a fraction. Therefore a wave winding is not possible. However a wave winding can

be made possible by considering one of the coils as dummy. Therefore number of active coils =

81 and number of commutator segments = 81.

Outside width of one segment and mica = Commutator segment pitch

=segments of number

DCπ

= thatassumption e with th81

25 x0.7xπ

DC = 0.7 D

= 0.68 cm

Maximum thickness of the brush = 5 times the commutator segment pitch

= 5 x 0.68 = 3.4 cm

Let the thickness of the brush tb = 2.5 cm

Armature current Ia = A66.3 250

0.9 /746 x20

V

/746 xHp==

η

Cross-sectional area of the brush / spindle Ab =b

a

A

I2

δ =

6 x2

66.3 x2= 11.65 cm

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Let the standard brush width Wb = 1.6 cmNumber of brushes / spindle nb =

bb

b

Wt

A =

1.6 x2.5

11.65 = 2.76 and is not possible

= 3 (say)

Length of the commutator LC = (1.6 + 0.5) 3 + 2 + 2 + 2 = 12.3 cm

Brush contact loss = Voltage drop / brush set x Ia = 2.0 x 66.3 = 132.6W

Example.3

A 600 kW, 6 pole lap connected D.C. generator with commutating poles running at 1200 rpm

develops 230V on open circuit and 250V on full load. Find the diameter of the commutator,

average volt / conductor, the number of commutator segments, length of commutator and brushcontact loss. Take Armature diameter = 56 cm, number of armature conductors = 300, number

of slots = 75, brush contact drop = 2.3 V, number of carbon brushes = 8 each 3.2 cm x 2.5 cm.

The voltage between commutator segments should not exceed 15V.

[ Note :

1. The D.C. generator is a cumulative compound one, with 230V on open circuit and 250V on

full load. Therefore while calculating the load current, 250V is to be considered.



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2. The number of commutator segments or coils and hence the number of turns / coil must be

so selected that the voltage per segment is not greater than 15V.

3. For a given voltage between segments, the volt / conductor goes on reducing as the number

of turns / coil goes on increasing. Thus the volt / conductor is maximum when the turns / coil is

minimum or turns / coil is one.

4.

Volt / conductor = (voltage between segments) / (conductors /coil) or (2 x number of turnsper coil)

5. There are 8 brushes / spindle of width 3.2 cm or 2.5 cm. ]

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Chapter.3 Design of Commutator and Brushes